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Calcium- 47 is a \(\beta^{-}\) emitter with a half-life of 4.5 days. If a bone sample contains \(2.24 \mathrm{~g}\) of this isotope, at what rate will it decay?

Short Answer

Expert verified
The decay rate is approximately \(4.42 \times 10^{21}\) decays per day.

Step by step solution

01

Understand the Properties of the Isotope

Calcium-47 is a beta-minus emitter, which means it decays by emitting an electron and an antineutrino. The half-life of Calcium-47 is given as 4.5 days, which implies that every 4.5 days, half of the isotope undergoes decay.
02

Calculate the Decay Constant

The decay constant (\( \lambda \)) can be calculated using the half-life (\(t_{1/2}\)) with the formula: \[ \lambda = \frac{\ln(2)}{t_{1/2}} \] where \( t_{1/2} = 4.5 \text{ days} \). Thus, \( \lambda = \frac{0.693}{4.5} \approx 0.154 \text{ day}^{-1} \).
03

Determine the Moles of Calcium-47

To find the moles of Calcium-47, we need its molar mass. The molar mass of Calcium-47 is approximately 47 g/mol. Using the mass of the sample: \[ \text{Moles of } \text{Ca-47} = \frac{2.24 \text{ g}}{47 \text{ g/mol}} \approx 0.04766 \text{ mol} \].
04

Find the Number of Nuclei

Using Avogadro's number \( (6.022 \times 10^{23} \text{ nuclei/mol}) \), calculate the actual number of Calcium-47 nuclei: \[ N = 0.04766 \text{ mol} \times 6.022 \times 10^{23} \text{ nuclei/mol} \approx 2.87 \times 10^{22} \text{ nuclei} \].
05

Calculate the Decay Rate

The decay rate (\( R \)) is given by the product of the decay constant and the number of nuclei: \[ R = \lambda N = 0.154 \text{ day}^{-1} \times 2.87 \times 10^{22} \text{ nuclei} \approx 4.42 \times 10^{21} \text{ decays/day} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calcium-47
Calcium-47 is an isotope of the element calcium. It is of particular interest in nuclear chemistry and medical applications, such as diagnostic imaging. This specific isotope has a mass number of 47, which denotes the number of protons and neutrons in its nucleus. In the nucleus of Calcium-47, there are 20 protons, like all calcium isotopes, and 27 neutrons. Calcium naturally occurs as a mix of stable isotopes with Calcium-47 differing in that it is radioactive. This radioactivity is characterized by its unstable nucleus, which undergoes decay to achieve a more stable state. Understanding the behavior and properties of such isotopes is crucial for fields like radiography and therapy planning. Examining the decay pattern and half-life of isotopes like Calcium-47 allows scientists and medical professionals to harness their potential while considering safety impacts.
Beta-minus emission
Beta-minus emission is a type of radioactive decay where a neutron in the nucleus of an unstable atom converts into a proton. This process emits an electron (known as a beta particle) and an antineutrino. In the context of Calcium-47, beta-minus decay is the mechanism by which it transforms into a different element, eventually stabilizing as it emits radiation. This form of decay increases the atomic number by one but keeps the mass number unchanged. In simple terms, during beta-minus decay, Calcium-47 will turn into a Scandium-47 atom. Understanding the details of beta-minus decay helps in comprehending how isotopes change over time and form the basis of nuclear reactions used in many applications.
Decay constant
The decay constant, represented by the symbol \( \lambda \), is a fundamental characteristic of radioactive decay processes. It quantifies the rate at which a radioactive isotope transitions into a stable state. Physically, it is the probability per unit time that a given nucleus will decay. For Calcium-47, the decay constant can be calculated using its half-life, with the formula for the decay constant being: \[ \lambda = \frac{\ln(2)}{t_{1/2}} \] where \( t_{1/2} \) is the half-life of the substance. Utilizing this formula provides a decay constant for Calcium-47 which approximately equals 0.154 days^{-1}. This value indicates how rapidly isotopes lose their radioactivity and can guide safe handling and usage in practical applications.
Half-life
The concept of half-life is crucial in understanding radioactivity and decay processes. The half-life is the time required for half of the radioactive nuclei in a sample to undergo decay. For Calcium-47, this time period is 4.5 days, which suggests that after each 4.5 days, only half of the original sample remains undecayed. Knowing the half-life is important because it helps predict how long a radioactive substance will remain active. It plays a significant role in determining the dosage for medical treatments using radioactive substances, ensuring both effectiveness and safety. Understanding half-life allows one to measure radioactive decay without continuously monitoring the substance. One simply needs to know the initial amount and the half-life to calculate how much will remain after a given period.
Radioactive decay calculations
Radioactive decay calculations allow scientists and engineers to predict how isotopes change over time. They involve determining quantities like decay rates and remaining amounts of substances, based on half-life and decay constant values. For Calcium-47, calculating the decay rate involves several steps:
  • Determine the molar quantity using mass and molar mass.
  • Calculate the number of nuclei with Avogadro's number.
  • Use the decay constant alongside the number of nuclei to find the decay rate.
The decay rate for a sample, as seen in the original example, is determined by multiplying the number of Calcium-47 nuclei by the decay constant. This results in the rate at which the isotope undergoes decay, denoted in units like decays per day. Proper understanding of these calculations is essential in fields utilizing radioactive substances to ensure accurate predictions and safe application of radiological principles.

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Most popular questions from this chapter

Show that the net result of the proton-proton fusion chain that occurs inside our sun can be summarized as $$ 6 \mathrm{p}^{+} \rightarrow{ }_{2}^{4} \mathrm{He}+2 \mathrm{p}^{+}+2 \beta^{+}+2 \gamma+2 \nu_{\mathrm{e}} $$

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An unstable isotope of cobalt, \({ }^{60} \mathrm{Co}\), has one more neutron in its nucleus than the stable \({ }^{59} \mathrm{Co}\) and is a beta emitter with a half-life of 5.3 years. This isotope is widely used in medicine. A certain radiation source in a hospital contains \(0.0400 \mathrm{~g}\) of \({ }^{60} \mathrm{Co}\). (a) What is the decay constant for that isotope? (b) How many atoms are in the source? (c) How many decays occur per second? (d) What is the activity of the source, in curies?

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